r + s = 4 and 2r + 3s = 8 multiply the first equation by 2 giving 2r + 2s = 8 subtract this from the second equation giving s = 0 So r = 4 and s = 0.
The sum = 2r : where r is the row number.
12-2r = 10
2r-1=9 2r =10 r =5
11
2r + 2s = 50 2r - s = 17 therefore 4r - 2s = 34 Add so that you can eliminate one of the variables: 2r + 2s = 50 4r - 2s = 34 ---------------- 6r + 0s = 84 Solve for r: 6r = 84 r = 14 Substitute r into one of the original equations: 2(14) + 2s = 50 28 + 2s = 50 2s = 22 s = 11 Doublecheck with the other original equation: 2(14) - 11 = 28 - 11 = 17
r + s = 4 and 2r + 3s = 8 multiply the first equation by 2 giving 2r + 2s = 8 subtract this from the second equation giving s = 0 So r = 4 and s = 0.
r - 10 + 100 - 3r = -2r + 90
The difference between the number of 2s and the number of 4s in a sum.
The sum = 2r : where r is the row number.
|2r + 9| = 5 So 2r + 9 = 5 or 2r + 9 = -5 so that 2r = -4 or 2r = -14 r = -2 or r = -7
The equation 4R + 3s + 2r = 6r + 3s is an example of the distributive property of addition, where the term 4R is being distributed over the sum of 2r and 6r. To see this more clearly, we can rewrite the equation as: 4R + 3s + 2r = (4R + 6r) + 3s Notice how the terms 4R and 6r are combined and the distributive property allows us to simplify the left-hand side of the equation.
12-2r = 10
4r2-25 = (2r)2-52 = (2r-5)(2r+5)
2r-1=9 2r =10 r =5
11
2r + 5 - - 1 When simplified: 2r + 6